The "weak charged current" is associated with a particle called the $W$, which can have either electrical charge, usually written $W^\pm$. Particles which interact with the charged current, which are the six quarks and six leptons of the standard model, can be thought of as different states of their partners in a direction called "isospin space," and the $W^\pm$ is the operator which causes rotations between one particle flavor and its partner.
That is to say, there is a very real way in which the electron and the electron-flavored neutrino are different states of the same particle, and transitions like
$$\mathrm e^- \longleftrightarrow W^-\nu_\mathrm e \quad\text{or}\quad \mathrm e^-W^+\longleftrightarrow\nu_\mathrm e$$
are allowed to happen just all the time, unless there is some other rule which prevents it. The most common reason these flavor oscillations don't happen is energy conservation. The $W$ is very heavy, so there has to be an awful lot of energy kicking around to end up with a real $W$ particle in the final state.
In a reaction where you have
$$
\text{(stuff)} + \mathrm p + \mathrm e^- \stackrel?\longrightarrow{}
\text{(stuff)} + \mathrm n + \nu_\mathrm e
$$
it may or may not be the case that the total mass-energy of the initial state on the left can be more than the total mass-energy of the final state on the right. If the (stuff) is the vacuum, there's not enough energy to proceed unless the electron and proton both have a lot of kinetic energy; therefore hydrogen atoms can't decay by electron capture. But if the $(\text{stuff}+\mathrm p)$ is a potassium-40 nucleus, then $(\text{stuff}+\mathrm n)$ is an argon-40 nucleus. You might imagine that argon-40 is more tightly bound than potassium-40 because there are fewer positively-charged protons in the nucleus to mutually repel each other. (Homework assignment: look up all the decays of potassium-40 and convince yourself electrical repulsion isn't a great argument for this particular mass number.)
If the decay is allowed energetically, you can have the two isospin rotations
$$
\mathrm p \to W^+ \mathrm n \quad\text{and}\quad W^+ \rm e^- \to \nu_e
$$
using the same $W^+$. (You can also imagine a $W^-$ if you put the electron transition "first.") Because the $W$ doesn't exist in the final state, we actually don't need the decay to have enough energy to create it at all. We can say instead that the decay is mediated by a "virtual $W$," or even that the decay is mediated by "the charged weak current."
If you want to compute the rate at which these decays occur, you have to know something about how strongly the weak interaction couples to the different particles involved. Those coupling constants are feeble (thus the name "weak interaction"), so weak decays tend to be much slower than electromagnetic or strong decays.